I had thought that if under an infinitesimal rotation by [tex]\theta[/tex] around the(adsbygoogle = window.adsbygoogle || []).push({}); z-axis a quantity transforms according to

[tex]\psi\rightarrow\left(1+i\left[S+\left(\frac{1}{i}x\frac{\partial}{\partial y}-\frac{1}{i}y\frac{\partial}{\partial x}\right)\right]\theta\right)\psi[/tex]

then the operator S is by definition thez-component of the spin operator for that quantity.

However, the Dirac field operator transforms (in the appropriate representation) according to

[tex]\psi\rightarrow\left(1+i\left[\frac{1}{2}\left(\begin{array}{cc}\sigma_{z}&0\\ 0&\sigma_{z}\end{array}\right)+\left(\frac{1}{i}x\frac{\partial}{\partial y}-\frac{1}{i}y\frac{\partial}{\partial x}\right)\right)\theta\right]\right)\psi[/tex]

yet https://www.amazon.com/gp/product/0521478146/103-7918027-3072619?v=glance&n=283155 states that [tex]\tau_{z}=\frac{1}{2}\left(\begin{array}{cc}\sigma_{z}&0\\0&\sigma_{z}\end{array}\right)[/tex] cannot be the spin operator because it does not commute with [tex]\gamma^{\mu}\partial_{\mu}[/tex] The correct spin operator is--or has something to do with--the Pauli-Lubanski pseudovector.

My question is, first, what's wrong with not commuting with [tex]\gamma^{\mu}\partial_{\mu}[/tex]? Why is that necessary for the a good spin operator?

And secondly, if [tex]\tau_{z}[/tex] is not a good spin operator, what about the fact that it generates the non-scalar part of a rotation? Is that not the definition of spin?

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# What is spin? (advanced question)

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